Approximation algorithms for k-echelon extensions of the one warehouse multi-retailer problem

Abstract

In this paper, we consider k-echelon extensions of the deterministic one warehouse multi-retailer problem. We give constant factor approximation algorithms for some of these extensions when k is fixed. We focus first on the case without backorders and we give a $(2k-1)$-approximation algorithm under general assumptions on the evolution of the holding costs as products move toward the final customers. We then improve this result to a k-approximation when the holding costs are monotonically non-increasing or non-decreasing (which is a natural situation in practice). Finally we address problems with backorders: we give a 3-approximation for the one-warehouse multi-retailer problem with backlog and a k-approximation algorithm for the k-level Joint Replenishment Problem with backlog (a variant where inventory can only be kept at the final retailers). Ours results are the first constant approximation algorithms for those problems. In addition, we demonstrate the potential of our approach on a practical case. Our preliminary experiments show that the average optimality gap is around 15%.

Appendices

Appendix A: Proof of Lemma 1

Given an optimal solution to OPT, i.e. a set of orders at each location and a set of paths to flow each demand $d_{jt}$ for all retailers j, we want to produce (feasible) solutions to the problems $(S_i)$ whose total cost does not exceed the cost of OPT. We consider solutions obtained by keeping the same set of orders at location i. So the total ordering costs will be the same in OPT and in the feasible solutions we propose for the problems $(S_i)$’s. Now we need to argue about the holding cost of flowing the units. Consider a retailer j, it is involved in k problems $(S_j)$, $(S_{p(j)})$, ..., $(S_{p^{k-1}(j)})$. If we show that the holding cost for product j in those problems is bounded above by $\frac{1}{k}$ times the holding cost for product j in OPT, then it proves that the total cost of the solutions we provide is a lower bound on OPT. Moreover as $\mathcal H_i + \mathcal K_i$ is the cost of an optimal solution to $(S_i)$, it follows that the sum of those costs is a lower bound on OPT too (we can only improve upon our feasible solutions by considering the optimal ones).

So let us consider retailer j and one of its ancestor $i=p^\ell (j)$ for some $\ell \in \{1,\ldots ,k-1\}$. We will argue on each flow of demand separately. Let $\mathcal P_{jt} = (s_0,\ldots ,s_{k-1})$ be the path along which we flow demand $d_{jt}$ in OPT. In $(S_i)$ we will serve demand for product j in period t by the order $s_\ell$ at location i. The holding cost for flowing the demand $d_{jt}$ from $s_\ell$ to t in OPT is greater or equal to $d_{jt} \cdot h'_j \cdot (t-s_\ell )$ as $h'_j=\min _{0\le \bar{\ell } \le \ell } h_{p^{\bar{\ell }}(j)}$ and hence the corresponding holding cost in S(i) ($=\frac{1}{k} d_{jt} \cdot h'_j \cdot (t-s_\ell )$ as we consider only the demand $\frac{1}{k} d_{jt}$) is bounded above by $\frac{1}{k}$ times the holding cost for those units in OPT. The result follows. $\square$

Appendix B: Proof of Lemma 2

Let us prove this by induction on the value of k. If $k=1$ then there is nothing to prove. So let us assume the property is true for a system with $k-1$ levels and we will prove it for a system with k levels. Let us consider the path at latest ${\mathcal P}_{it}(\mathcal O) \sim (s_0,s_1,\ldots ,s_{k-1})$ for a demand $d_{it}$ (possibly $d_{it}=0$).

We first prove property (i).

Suppose first that $s_0 \ne s_1$. There is no order at p(i) in the interval $(s_1,s_0]$ otherwise $s_1$ would not be defined properly. There is also no order in $(s_0,t]$ at p(i) because otherwise, let $t'$ be the time period of the first one, the procedure uncrossRight($\mathcal O_i$,$\bar{\mathcal O}_{1}$) would have created an extra order in time period $t'$ at location i (indeed let s be the first order of $\mathcal O_i$ in $(s_1,s_0]$. s exists as $s_0\in \mathcal O_i$ and Algorithm 2 examines s and then an order is created in $t'$), contradicting the definition of $s_0$. Hence there is no order in $(s_1,t]$ at location p(i). Thus, by induction, the property holds for the system restricted to the $k-1$ first levels (i.e. if we remove level k and consider $\mathcal P_{p(i),t}({\mathcal O}) \sim (s_1,\ldots ,s_{k-1})$) and this extends for the system with k levels as the only thing that remains to be shown is that there is no order at p(i) in the interval $(s_1,t]$, which we already mentioned.

Fig. 11

Results for 10 random demand profiles

Suppose now that $s_0 = s_1$. Let us define $j=\min \{j'=1,\ldots ,k: s_{j'} \ne s_0\}$ (with the convention $s_k=-1$). If $j=k$ then property (i) trivially holds. Now if $j\ne k$, we want to prove that there is no order in $(s_0,t]$ at location $p^j(i)$. Indeed if this later fact is true, the property holds by induction following the same reasonning as above. So assume the contrary, and let $t'$ be the first time period in $(s_0,t]$ for which there is an order at location $p^j(i)$. Because $s_{j-1} \ne s_j$, the procedure uncrossRight($\mathcal O_{p^{j-1}(i)}$,$\bar{\mathcal O}_{1}$) would have created an extra order in time period $t'$ at location $p^{j-1}(i)$, if it would not already exist (see above). Then similarly, the procedure uncrossRight($\mathcal O_{p^{j-2}(i)}$,$\bar{\mathcal O}_{2}$) would have created an extra order in time period $t'$ at location $p^{j-2}(i)$, if it would not already exist, and so on. Hence there would be an order in period $t'$ at location i, a contradiction with the definition of $s_0$.

Observation: Showing that the procedure uncrossRight($\mathcal O_{p^{j-2}(i)}$,$\bar{\mathcal O}_{2}$) would have created an extra order in time period $t'$ at location $p^{j-2}(i)$ is less straightforward because $s_{j-2}$ need not be in $\mathcal O_{p^{j-2}(i)}$ before uncrossing. However one can prove that there is an order of $\mathcal O_{p^{j-2}(i)}$ in $(t'',t')$ where $t''$ is the previous period where there is a direct link from $p^{j}(i)$ to $p^{j-1}(i)$ – where $t''=0$ by convention if there is none. Indeed, if not, then there would be no order in period $s_{j-2}$ at $p^{j-2}(i)$. Then we can argue similarly as above. The argument extends to $s_{j-3}$ and so on.

We prove now property (ii).

The property is true up to level $k-1$ by induction (if we remove level k and consider $\mathcal P_{p(i),s_0}({\mathcal O})\sim (s_1,\ldots ,s_{k-1}$) i.e. for all $j=2,\ldots ,k-1$, for all $\ell =2,\ldots ,j$ there exists an order in period $s_j$ at location $p^{\ell -1}(i)$. It thus remains to prove that for all $j=1,\ldots ,k-1$ there exists an order in period $s_j$ at location i. Suppose there is no order at location i in period $s_j$ for some j in $\{1,\ldots ,k-1\}$ and assume that j is the minimum $j' \ge 1$ such $s_j=s_{j'}$. We have by construction $s_0 \ge s_{j-1} > s_j$ and there is a direct link from $p^{j}(i)$ to p(i) by induction. But then the procedure uncrossLeft($\mathcal O_i$,$\bar{\mathcal O}_{j}$) would have created an extra order in period $s_j$ at location i, a contradiction (here again we can argue similarly to the above Observation that there is an order of $\mathcal O_i$ in $(s_j,t'')$ where $t''$ is the next period for which there is a direct link from $p^{j}(i)$ to p(i). Note that $t'' > s_0$.).

Operations ($\star$) add at most $k-1$ orders at location i for each existing one by Claim 1. The same holds for operations $(\sharp )$. The result follows (Fig. 11). $\square$